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template.tex
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template.tex
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\documentclass
[kulak,outline,totalframes,UKenglish]
%%% OPTIONS %%%
% kulak / kul changes logo
% outline adds an outline frame at the start of each section
% totalframes adds the total number of frames to the footer
% handout handout mode, use this when printing slides
% <language> e.g. dutch, UKenglish (passed on to babel)
{kulakbeamer}
%%% PACKAGES %%%
% No need to load: babel, inputenc, fontenc, lmodern, tikz, xcolor, keyval, geometry, hyperref
\usepackage{mathtools,amsthm,amssymb}
\usepackage{tikz-cd}
%%% DATA %%%
\title[Short presentation title]{Actual full-length title which may be very long and not fit on a single line}
%\subtitle{Subtitle}
\author[Short Name]{Long Name}
\institute[Kulak]{KU Leuven Kulak Kortrijk Campus}
\date{29th April 1967}
% Uncomment and edit this when using a different language!
%\renewcommand{\outlinename}{Overzicht}
\begin{document}
%%% SLIDES %%%
\section{Introduction}
\begin{frame}{Introduction}
Welcome to my presentation!
\end{frame}
\section[Short section title]{Long section title}
\subsection{Subsection title}
\begin{frame}{Frame title}
Frame content.
\end{frame}
\subsection{AMS-LaTeX}
\begin{frame}{Lefschetz fixed point theorem}
\begin{theorem}[S. Lefschetz]
Let \(f \colon X \to X\) be a continuous self-map on a connected, compact polyhedron \(X\).
If \(L(f) \neq 0\), then \(f\) has at least one fixed point.
\end{theorem}
\medskip
The converse to this theorem need not be true.
\medskip
\begin{example}
The identity map \(\operatorname{id}_{S^1}\) on the circle \(S^1\) has Lefschetz number \(L(\operatorname{id}_{S^1}) = \chi(S^1) = 0\), but obviously it has infinitely many points.
\end{example}
\end{frame}
\subsection{Diagrams}
\begin{frame}[fragile]{General lifting lemma}
Let \(p \colon Z \to Y\) be a covering map and fix \(y \in Y\) and \(z \in Z\) such that \(p(z) = y\).
Let \(f \colon X \to Y\) be a continuous map with \(f(x) = y\).
Suppose that \(Z\) is path-connected and locally path-connected.
\medskip
If
\begin{equation*}
f_\pi ( \pi_1(X,x) ) \subseteq p_\pi(\pi_1(Z,z)),
\end{equation*}
then there exists a continuous map \(\tilde{f} \colon X \to Z\) such that:
\begin{itemize}
\item \(p \circ \tilde{f} = f\), i.e.\@ the diagram below commutes,
\item \(\tilde{f}(x) = z\).
\end{itemize}
\[
% Requires the frame to be given the "fragile" option
\begin{tikzcd}
& Z \arrow[d,"p"]\\
X \arrow[ur,"\tilde{f}"] \arrow[r,"f"] & Y
\end{tikzcd}
\]
\end{frame}
\subsection{Colours}
\begin{frame}{Main Colours}
\bigskip
\Large
\begin{tabular}{lll}
Primary Blue & KULblue1 & {\textcolor{KULblue1}{\rule{1cm}{1cm}}}\\
Secondary Blue & KULblue2 & {\textcolor{KULblue2}{\rule{1cm}{1cm}}}\\
\end{tabular}
\end{frame}
\begin{frame}{Accent Colours}
\bigskip
\begin{tabular}{ll@{\hskip 60pt}ll@{\hskip 60pt}ll}
KULblue3a & {\textcolor{KULblue3a}{\rule{0.5cm}{0.5cm}}}&
KULblue3b & {\textcolor{KULblue3b}{\rule{0.5cm}{0.5cm}}}&
KULblue3c & {\textcolor{KULblue3c}{\rule{0.5cm}{0.5cm}}}\\&&&&&\\
KULcyan & {\textcolor{KULcyan}{\rule{0.5cm}{0.5cm}}}&
KULorange & {\textcolor{KULorange}{\rule{0.5cm}{0.5cm}}}&
KULdarkgreen & {\textcolor{KULdarkgreen}{\rule{0.5cm}{0.5cm}}}\\&&&&&\\
KULyellow & {\textcolor{KULyellow}{\rule{0.5cm}{0.5cm}}}&
KULgreyblue & {\textcolor{KULgreyblue}{\rule{0.5cm}{0.5cm}}}&
KULbrown & {\textcolor{KULbrown}{\rule{0.5cm}{0.5cm}}}\\&&&&&\\
KULred & {\textcolor{KULred}{\rule{0.5cm}{0.5cm}}}&
KULgold & {\textcolor{KULgold}{\rule{0.5cm}{0.5cm}}}&
KULpurple & {\textcolor{KULpurple}{\rule{0.5cm}{0.5cm}}}\\&&&&&\\
KULpink & {\textcolor{KULpink}{\rule{0.5cm}{0.5cm}}}&
KULlightgreen & {\textcolor{KULlightgreen}{\rule{0.5cm}{0.5cm}}}&
KULredbrown & {\textcolor{KULredbrown}{\rule{0.5cm}{0.5cm}}}\\
\end{tabular}
\end{frame}
\section{Conclusion}
\begin{frame}{Closing frame}
Make sure to thank your audience and ask if there are any questions.
\end{frame}
\end{document}